Asked Nov 7, 2019

When 6.78 g of solid ammonium nitrate dissolves in 50.0 g of water at 18.9oC, the temperature of the solution drops to 4.0oC. What is q for the reaction?


Expert Answer

Step 1


Specific heat of water = 4.18 J/g. oC

Mass of ring = 50 g

Mass of ammonium nitrate = 6.78 g.

Initial Temperature of water = 18.9 oC

Final temperature of water = 4.00 °C

Step 2

Heat loss or gain i...


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q = mCAT


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Chemical Thermodynamics

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